I have a question on a paper of Azad and Kobayashi "Ricci-flat Kähler metrics on symmetric varieties". Here is the link: http://www.academia.edu/2579043/Ricci-flat_Kahler_metrics_on_symmetric_varieties

On the first, there is mentioned the main result: "Theorem (1.1): Let $G^{\mathbb{C}} / K^{\mathbb{C}}$ be a symmetric variety. Then there exists a $G$-invariant complete Ricci-flat Kähler metric on $G^{\mathbb{C}} / K^{\mathbb{C}}$." Further below in section (1.2) (1) I do not understand the statement: "If $\sqrt{-1}\partial \overline{\partial} P$ is a Ricci-flat complete Kähler metric in Theorem (1.1), then $\bigwedge^{top}(\sqrt{-1} \partial \overline{\partial} P) = \eta \wedge \overline{\eta}$.", where $\eta$ is the $G^{\mathbb{C}}$-invariant top degree holomorphic volume form on $G^{\mathbb{C}} / K^{\mathbb{C}}$. I do not understant how one obtains the equation $\bigwedge^{top}(\sqrt{-1} \partial \overline{\partial} P) = \eta \wedge \overline{\eta}$ from the above assumptions?

I have to admit that I am not very experienced in this field and right now I am in the state of learning. Hope that some of you could help me out with this question.

Greetings. Bernard

  • $\begingroup$ Both of the forms you list are nonzero $G^\mathbb{C}$-invariant top degree forms, so they have to be constant multiples of one another for that reason alone. That said, they can't be equal when the dimension $n$ of $G^\mathbb{C}/K^\mathbb{C}$ is odd because the top exterior power of the Kähler form is real, while $\eta\wedge\overline{\eta}$ is purely imaginary when $n$ is odd. Thus, the authors have been a little careless. There should at least have been a normalizing factor in front of the $\eta\wedge\overline{\eta}$ to make it real. Otherwise, this is really a question of normalizing $\eta$. $\endgroup$ – Robert Bryant Apr 11 '13 at 12:45
  • $\begingroup$ Yes but I thaught that the Kähler metric constructed in this paper is only $G$-invariant, not $G^{\mathbb{C}}$, as it is mentioned in Theorem (1.1). Does it still work for such a $G$-invariant metric? $\endgroup$ – bernard Apr 11 '13 at 12:52
  • $\begingroup$ I see that you are right; I had been careless and read $G^\mathbb{C}$ where you had only written $G$, so my argument doesn't apply. Looking at the source that you cite, I see that in (1.2) the authors are only claiming that they are going to show this later, not claiming that it is obvious at that point. Thus, you should be patient and continue reading the paper; presumably, it will be explained when the prove their theorem later on. (I haven't read through to check this, I admit.) $\endgroup$ – Robert Bryant Apr 11 '13 at 14:51

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